(tanα-3)(sinα+cosα+3)=0
sinα+cosα+3<>0，tanα-3=0，tanα=3。
(4cosα-2sinα)/(5cosα+3sinα)=(4-2tanα)/(5+3tanα)=(4-6)/(5+9)=-1/7
(2/3)(sinα)^2+(1/4)(cosα)^2
=[(2/3)(sinα)^2+(1/4)(cosα)^2]/[(sinα)^2+(cosα)^2]
=[(2/3)(tanα)^2+1/4]/[(tanα)^2+1]
=5/8
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